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I am going through this document to read up a bit about functional calculus. I have a question on the proof for the variation of the nonlocal functional, given on page 6. The nonlocal functional is defined as $$F_w[f] = \int _{a}^b dx_1 \int _{a}^b dx_2 w(x_1, x_2) f(x_1) f(x_2)$$ And the author states that $$ \frac{\delta F_w}{\delta f(x_1)} = \int dx_2 f(x_2)[w(x,x_2)+w(x_2,x)]$$

I do not quite understand how this was achieved. The following is my proof for it:

$$F_w[f+\epsilon \eta] - F_w [f] = \delta F_w = \int dx_1 \int dx_2 w(x_1, x_2)[\epsilon f(x_1) \eta (x_2) + \epsilon f(x_2) \eta (x_1) + \epsilon ^2 \eta (x_1) \eta (x_2)$$ In this document, the functional derivative of $F_w$ has been defined as $$ \frac{dF_w [f+\epsilon \eta]}{d\epsilon} \Big| _{\epsilon = 0} = \int dx_1 \frac{\delta F_w}{\delta f(x_1)} \eta (x_1) $$

Doing some manipulations and ordering terms...

$$ \frac{dF_w [f+\epsilon \eta]}{d\epsilon} \Big| _{\epsilon = 0} = \int dx_1 \eta (x_1) \left[ \int dx_2 w(x_1, x_2) f (x_2) \right]+ \int dx_1 \eta (x_1) \left[\int dx_2 w(x_1, x_2) \frac{\eta (x_2)}{\eta (x_1)} f(x_1)\right] $$

Everything inside the big square brackets "[]" should feature in $\frac{\delta F_w}{\delta f(x_1)}$. However, I don't understand how I can reasonable scratch out the pesky $\eta(x_2)/\eta (x_1)$ term.

Is my reasoning valid? Am I messing up the computation somewhere? Any advice you have is appreciated!

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You didn't compute correctly the derivative with respect to $\varepsilon$; indeed, we have : $$ \begin{array}{rcl} \displaystyle\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\right|_{\varepsilon=0}F_w[f+\varepsilon\eta] &=& \displaystyle \lim_{\varepsilon\rightarrow0}\frac{F_w[f+\varepsilon\eta]-F_w[f]}{\varepsilon} \\ &=& \displaystyle \lim_{\varepsilon\rightarrow0} \int\mathrm{d}x_1\int\mathrm{d}x_2\, w(x_1,x_2)\left(f(x_1)\eta(x_2) + f(x_2)\eta(x_1) + \varepsilon\eta(x_1)\eta(x_2)\right) \\ &=& \displaystyle \int\mathrm{d}x_1\int\mathrm{d}x_2\, w(x_1,x_2)\left(f(x_1)\eta(x_2)+f(x_2)\eta(x_1)\right) \\ &=& \displaystyle \int\mathrm{d}x_1\, \eta(x_1)\left(\int\mathrm{d}x_2\, w(x_1,x_2)f(x_2)\right) \;+\\&& \displaystyle \int\mathrm{d}x_2\, \eta(x_2)\left(\int\mathrm{d}x_1\, w(x_1,x_2)f(x_1)\right) \\ &=& \displaystyle \int\mathrm{d}x_1\, \eta(x_1)\left(\int\mathrm{d}x_2\, (w(x_1,x_2)+w(x_2,x_1))f(x_2)\right) \end{array} $$ by exchanging the variable names in the second integral, i.e. $x_1 \leftrightarrow x_2$, hence $$ \frac{\delta F_w[f]}{\delta f(x_1)} = \int\mathrm{d}x_2\, (w(x_1,x_2)+w(x_2,x_1))f(x_2) $$ by comparison with $\displaystyle \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\right|_{\varepsilon=0}F_w[f+\varepsilon\eta] = \int\mathrm{d}x_1\,\eta(x_1)\, \frac{\delta F_w[f]}{\delta f(x_1)}$.


Addendum. I see that your document talks about applications of variational calculus in physics. Note that this way of computing functional derivatives mainly serves as a definition (to please mathematicians =P). In practice, the function $\eta(x)$ is replaced by the "functional infinitesimal" $\delta f(x)$, such that $\frac{\delta f(x')}{\delta f(x)} = \delta(x'-x)$, where $\delta$ is the Dirac delta function, which permits to write the functional derivative as follows : $$ \frac{\delta F[y(x')]}{\delta y(x)} = \delta(x'-x)\sum_{n=0}^\infty (-1)^n\frac{\mathrm{d}^n}{\mathrm{d}x^n} \frac{\partial^nF}{\partial y^{(n)}}, $$ which corresponds to the well-known Euler-Lagrange operator $\frac{\partial}{\partial y} - \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial}{\partial y'}$ when only first derivatives appear in $F$. See this answer of mine for more details.

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  • $\begingroup$ This is great, thank you @Abezhiko! yes, the quick symmetric switch of variables makes sense $\endgroup$
    – megamence
    Commented Jan 22, 2023 at 15:58

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